Đáp án:

 Bài `3`

`a, x^2 - 2xy + y^2 - 9`

`= (x^2 - 2xy + y^2) - 9`

`= (x - y)^2 - 3^2`

`= (x - y - 3)(x - y + 3)`

`b, x^2 + 4x + 4 - y^2`

`= (x^2 + 4x + 4) - y^2`

`= (x^2 + 2.x.2 + 2^2) - y^2`

`= (x + 2)^2 - y^2`

`= (x + 2 + y)(x + 2 - y)`

`c, x^2 - 4y^2 + 3x - 6y`

`= [x^2 - (2y)^2] + (3x - 6y)`

`= (x - 2y)(x + 2y) + 3(x - 2y)`

`= (x - 2y)(x + 2y + 3)`

`d, x^3 - y^3 + 2x^2 - 2y^2`

`= (x^3 - y^3) + 2(x^2 - y^2)`

`= (x - y)(x^2 + xy + y^2) + 2(x - y)(x + y)`

`= (x - y)[x^2 + xy + y^2 + 2(x +y)]`

`= (x - y)(x^2 + xy + y^2 +2x + 2y)`

Bài `4`

`a, (x + 1)(x - 2) = (x + 1)^2`

`⇔ x^2 - 2x + x - 2 = x^2 + 2x + 1`

`⇔ x^2 - 2x + x - 2 - x^2 - 2x - 1 = 0`

`⇔ (x^2 - x^2) + (-2x + x - 2x) + (-2 - 1) = 0`

`⇔ - 3x - 3 = 0`

`⇔ - 3x = 3`

`⇔ x = -1`

Vậy `x = -1`

`b, x^2 + 4x + 3 = 0`

`⇔ x^2 + x + 3x + 3 = 0`

`⇔ x(x + 1) + 3(x + 1) = 0`

`⇔ (x + 1)(x + 3) = 0`

`⇒` $\left[\begin{matrix} x+1=0\\ x+ 3=0\end{matrix}\right.$

`⇒` $\left[\begin{matrix} x = -1\\ x = -3\end{matrix}\right.$

Vậy `x = -1` hoặc `x = -3`

`c, x^2 - 5x - 6 = 0`

`⇔ x^2 + x - 6x - 6 = 0`

`⇔ x(x + 1) - 6(x + 1) = 0`

`⇔ (x + 1)(x - 6) = 0`

`⇒` $\left[\begin{matrix} x+1=0\\ x - 6=0\end{matrix}\right.$

`⇒` $\left[\begin{matrix} x = -1\\ x = 6\end{matrix}\right.$

Vậy `x = -1` hoặc `x =6`

`d, x^3 - 2x^2 - x + 2 = 0`

`⇔ x^2 (x - 2) - (x - 2) =0`

`⇔ (x - 2)(x^2 - 1) = 0`

`⇔ (x - 2)(x - 1)(x + 1) = 0`

`⇒` $\left[\begin{matrix} x-2=0\\ x-1=0\\x + 1 = 0\end{matrix}\right.$

`⇒` $\left[\begin{matrix} x=2\\ x=1\\x = -1\end{matrix}\right.$

Vậy `x = 2, x = 1` hoặc `x = -1`

B3: c,$x^{2}$ -4$y^{2}$ +3x-6y                d,$x^{3}$-$y^{3}$+$2x^{2}$-$2y^{2}$ 

    =(x-2y)(x+2y)+3(x-2y)                         =($x^{3}$-$y^{3}$)+2($x^{2}$-$y^{2}$) 

    = (x-2y)[(x+2y)+3]                               =(x-y)($x^{2}$+xy+$y^{2}$ )+2(x-y)(x+y)             

    = (x-2y)(x+2y+3)                                 =(x-y)[($x^{2}$+xy+$y^{2}$)+2(x+y)]

B4: c, $x^{2}$ -5x-6=0

⇒ ($x^{2}$-5x+$\frac{25}{4}$)-$\frac{49}{4}$ =0

⇒(x-$\frac{5}{2}$)²-($\frac{7}{2}$)²=0

⇒( x-$\frac{5}{2}$+$\frac{7}{2}$ )(x-$\frac{5}{2}$-$\frac{7}{2}$)=0

⇒(x+1)(x-6)=0⇒x+1=0    hoặc x-6=0

*x+1=0                             *x-6=0

⇒x=-1                              ⇒x=6

Vậy x=-1 hoặc x=6

d,$x^{3}$-$2x^{2}$-x+2=0 

⇒$x^{2}$ (x-2)-(x-2)=0

⇒(x-2)($x^{2}$-1)=0

⇒(x-2)(x-1)(x+1)=0

⇒x∈{2;1;-1}